Question

Calculate pV2+ at each of the points in the titration of 21.17 mL of 0.0364 M...

Calculate pV2+ at each of the points in the titration of 21.17 mL of 0.0364 M EDTA with 0.0182 M VCl2 . The EDTA solution is buffered at a pH of 10.00 . The fraction of free EDTA in the Y4− form ( ?Y4− ) can be found in this table. The formation constant for the V2+−EDTA complex is given by log?f=12.7 .

4.234 mLpV2+=

16.94 mLpV2+=

33.87 mLpV2+=

41.49 mLpV2+=

42.26 mLpV2+=

42.34 mLpV2+=

42.42 mLpV2+=

46.57 mLpV2+=

50.81 mLpV2+=

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Answer #1

ANSWER:

Data:

  • EDTA = 0.0364 M and 21.17 mL
  • VCl2 = 0.0182 M
  • pH = 10.00
  • αY4- = 0.30 (at pH 10.00)
  • V2+-EDTA => log Kf = 12.7
    • Kf = 1012.7 = 5.01x1012

Titration reaction:

v+2+ EDTA +VY->

For the titration:

  • Before equivalence point: all V2+ added is transformed into VY-2. The [V+2] comes from dissociation of VY-2
  • Equivalence point: all V2+ added is transformed into VY-2. The moles of V+2 are equal to moles of EDTA and  from dissociation of VY-2
  • After equivalence point: excess of V2+. The [V+2] comes from unreacted V2+

Dissociation of VY-2:

v++ EDTA=VY- K,

VY- A IV+ EDT A a = Kjaya- = 0.30 x 5.01.21012 = 1.50x1012

The equivalence volume (VV+2) is:

mol +2 = mol EDTA

Vy+2 [V+2] = [EDT AVEDTA

Vy +2 = 1 [EDT AL X VEDTA [V+] 0.0364 M x 21.17 mL = 42.34 mL 0.0182 M

Before equivalence point:

  • The [V+2] comes from dissociation of VY-2

[v+2 = [VY-> vy- 5.985x10-14 x [EDTA K[EDT A

  • The concentration of VY-2 is

[VY-2-moles Vy- Vsel

  • All V2+ added is transformed into VY-2, then, the moles V2+= moles of VY-2

moles 1+2 Vy +2 + VEDTA v+2 x Vy+2 VV+2+ VEDTA

  • The concentration of EDTA is

unreacted moles EDTA LEDT A] = Vol

[EDTA] = initial moles EDT A - moles V+2 Vy+2 + VERTA

[EDTA] = [EDT A] VEDTA - [v+2] Vy+2 Vy+2 + VERTA

Vol V+2 [EDTA] [VY-2]
4.234 mL \frac{0.0364 \times 21.17-0.0182 \times 4.234}{21.17+4.234}=\mathbf{0.0273\, M} \frac{0.0182\times 4.234}{4.234+21.17}=\mathbf{3.03x10^{-3}}
16.94 mL \frac{0.0364 \times 21.17-0.0182 \times 16.94}{21.17+16.94}=\mathbf{0.0182\, M} \frac{0.0182\times 16.94}{16.94+21.17}=\mathbf{8.09x10^{-3}}
33.87 mL 0.0364 x 21.17 -0.0182 x 33.87 - = 0.0126 M 21.17 + 33.87 \frac{0.0182\times 33.87}{33.87+21.17}=\mathbf{1.12x10^{-2}}
41.49 mL \frac{0.0364 \times 21.17-0.0182 \times 41.49}{21.17+41.49}=\mathbf{0.0111\, M} \frac{0.0182\times 41.49}{41.49+21.17}=\mathbf{1.205x10^{-2}}
42.26 mL \frac{0.0364 \times 21.17-0.0182 \times 42.26}{21.17+42.26}=\mathbf{0.0109\, M} \frac{0.0182\times 42.26}{42.26+21.17}=\mathbf{1.21x10^{-2}}

then

Vol V+2 [V+2] pV+2
4.234 mL \frac{3.03x10^{-3}}{1.50x10^{12}\times 0.0273}=\boldsymbol{7.41x10^{-14}} -log\left ( 7.41x10^{-14} \right )=\mathbf{13.13}
16.94 mL \frac{8.09x10^{-3}}{1.50x10^{12}\times 0.0182}=\boldsymbol{2.96x10^{-13}} -log (2.96x10-13) = 12.53
33.87 mL \frac{1.12x10^{-2}}{1.50x10^{12}\times 0.0126}=\boldsymbol{5.93x10^{-13}} -log\left ( 5.93x10^{-13} \right )=\mathbf{12.23}
41.49 mL \frac{1.205x10^{-2}}{1.50x10^{12}\times 0.0111}=\boldsymbol{7.26x10^{-13}} -log\left ( 7.26x10^{-13} \right )=\mathbf{12.14}
42.26 mL \frac{1.21x10^{-2}}{1.50x10^{12}\times 0.0109}=\boldsymbol{7.39x10^{-13}} -log (7.39x10-13) = 12.13

At equivalence point (42.34 mL):

  • The concentration of VY-2 is calculated as before equivalence point

[VY^{-2}]=\frac{[V^{+2}]\times V_{V^{+2}}}{V_{V^{+2}}+V_{EDTA}}=\frac{0.0182\times 42.34}{42.34+21.17}=\mathbf{1.21x10^{-3}\, M}

  • At this point
V+2 + EDTA <--> VY-2
Initial 0 0 1.21x10-3
Final X X 1.21x10-3 - X

then

K_{f}'=\frac{[VY^{-2}]}{[V^{+2}][EDTA]}=\frac{1.21x10^{-3}-X}{X^{2}}=1.50x10^{12}

the great value of Kf allows this approximation:

\frac{1.21x10^{-3}}{X^{2}}=1.50x10^{12}

1.21.210-3 V 1.501012 = 2.84.r10-8 M

then

X=[V^{+2}]=\mathbf{2.84x10^{-8}\, M}

pV^{+2}=-log\left [V^{+2} \right ]=\mathbf{7.55}

After equivalence point:

  • The [V+2] comes from unreacted V2+

[V^{+2}]=\frac{unreacted\, moles\, V^{+2}}{V_{sol}}

[V^{+2}]=\frac{moles\, V^{+2}-initial\, moles\, EDTA}{V_{V^{+2}}+V_{EDTA}}

[V^{+2}]=\frac{[V^{+2}]\times V_{V^{+2}}-[EDTA]\times V_{EDTA}}{V_{V^{+2}}+V_{EDTA}}

then

Vol V+2 [V+2] pV+2
42.42 mL \frac{0.0182\times 42.42-0.0364\times 21.17}{42.42+21.17}=\mathbf{2.29x10^{-5}} -log\left ( 2.29x10^{-5} \right )=\mathbf{4.64}
46.57 mL \frac{0.0182\times 46.57-0.0364\times 21.17}{46.57+21.17}=\mathbf{1.14x10^{-3}} -log\left ( 1.14x10^{-3} \right )=\mathbf{2.94}
50.81 mL \frac{0.0182\times 50.81-0.0364\times 21.17}{50.81+21.17}=\mathbf{2.14x10^{-3}} -log\left ( 2.14x10^{-3} \right )=\mathbf{2.67}​​​​​​​
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